Count of Binary Operations on a Set

Theorem

Let $S$ be a set whose cardinality is $n$.

The number $N$ of possible different binary operations that can be applied to $S$ is given by:

$N = n^{\left({n^2}\right)}$

Proof

A binary operation on $S$ is by definition a mapping from the cartesian product $S \times S$ to the set $S$.

Thus we are looking to evaluate:

$N = \left|{\left\{{f: S \times S \to S}\right\}}\right|$

The domain of $f$ has $n^2$ elements, from Cardinality of Cartesian Product.

The result follows from Cardinality of Set of All Mappings.

$\blacksquare$

Comment

The number grows rapidly with $n$:

$\begin{array} {c|rr} n & n^2 & n^{\left({n^2}\right)}\\ \hline 1 & 1 & 1 \\ 2 & 4 & 16 \\ 3 & 9 & 19 \ 683 \\ 4 & 16 & 4 \ 294 \ 967 \ 296 \\ \end{array}$

We're still only at 4 elements in a set, and we're already up to 4 thousand million different possible algebraic structures.

This sequence is A002489 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): Exercise $4.1$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $2.3$